## Calculus: Early Transcendentals 8th Edition

A cubic function has one point of inflection. The x-coordinate of the inflection point is $~~\frac{x_1+x_2+x_3}{3}$
$y = ax^3+bx^2+cx+d$ $y' = 3ax^2+2bx+c$ $y'' = 6ax+2b$ We can find the value of $x$ such that $y'' = 0$: $y'' = 6ax+2b = 0$ $x = -\frac{b}{3a}$ $y$ has one point of inflection and the x-coordinate is $x = -\frac{b}{3a}$ Suppose $y = 0$ $y = ax^3+bx^2+cx+d = 0$ $x^3+\frac{b}{a}x^2+\frac{c}{a}x+\frac{d}{a} = 0$ Suppose $x_1, x_2,$ and $x_3$ are x-intercepts. $(x-x_1)(x-x_2)(x-x_3) = 0$ $(x^2-x_1x-x_2x+x_1x_2)(x-x_3) = 0$ $x^3-x_1x^2-x_2x^2+x_1x_2x-x_3x^2+x_1x_3x+x_2x_3x-x_1x_2x_3 = 0$ $x^3-x_1x^2-x_2x^2-x_3x^2+x_1x_2x+x_1x_3x+x_2x_3x-x_1x_2x_3 = 0$ $x^3-(x_1+x_2+x_3)x^2+(x_1x_2+x_1x_3+x_2x_3)x-x_1x_2x_3 = 0$ By comparing the two equations, we can see that $\frac{b}{a} = -(x_1+x_2+x_3)$ The x-coordinate of the inflection point is: $x = -\frac{b}{3a}$ $x = -\frac{1}{3}(\frac{b}{a})$ $x = -\frac{1}{3}[-(x_1+x_2+x_3)]$ $x = \frac{x_1+x_2+x_3}{3}$ The x-coordinate of the inflection point is $~~\frac{x_1+x_2+x_3}{3}$